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Python – Rearrange dictionary for consective value-keys
  • Last Updated : 10 May, 2020

Sometimes, while working with Python dictionaries, we can have a problem in which we need to rearrange dictionary keys so as a value is followed by same key in dictionary. This problem can have application in competitive programming algorithms and Graph problems. Let’s discuss certain ways in which this task can be performed.

Input : test_dict = {1 : 2, 3 : 2, 2 : 3}
Output : {1: 2, 2: 3, 3: 2}

Input : test_dict = {1 : 2}
Output : {1 : 2}

Method #1 : Using loop + keys()
The combination of above functions can be used to solve this problem. In this, we extract the dictionary keys using keys() and iterate till we find value succeded by equal key.

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# Python3 code to demonstrate working of 
# Rearrange dictionary for consective value-keys
# Using loop + keys()
  
# initializing dictionary
test_dict = {1 : 3, 4 : 5, 3 : 4, 5 : 6}
  
# printing original dictionary
print("The original dictionary : " + str(test_dict))
  
# Rearrange dictionary for consective value-keys
# Using loop + keys()
temp = list(test_dict.keys())[0]
res = {}
while len(test_dict) > len(res):
    res[temp] = temp = test_dict[temp]
  
# printing result 
print("The rearranged dictionary : " + str(res)) 

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Output :



The original dictionary : {1: 3, 4: 5, 3: 4, 5: 6}
The rearranged dictionary : {1: 3, 3: 4, 4: 5, 5: 6}

 

Method #2 : Using dictionary comprehension + accumulate()
The combination of above functions can be used to solve this problem. In this, we perform the task fo pairing using accumulate and rearrange new dictionary using dictionary comprehension.

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# Python3 code to demonstrate working of 
# Rearrange dictionary for consective value-keys
# Using dictionary comprehension + accumulate()
from itertools import accumulate
  
# initializing dictionary
test_dict = {1 : 3, 4 : 5, 3 : 4, 5 : 6}
  
# printing original dictionary
print("The original dictionary : " + str(test_dict))
  
# Rearrange dictionary for consective value-keys
# Using dictionary comprehension + accumulate()
res = {key : test_dict[key] for key in accumulate(test_dict, 
                              lambda key, x :test_dict[key])}
  
# printing result 
print("The rearranged dictionary : " + str(res)) 

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Output :

The original dictionary : {1: 3, 4: 5, 3: 4, 5: 6}
The rearranged dictionary : {1: 3, 3: 4, 4: 5, 5: 6}

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